Electric Field due to an Infinite Plan Sheet of Charge

Consider a thin, nonconducting, infinite plane sheet of charge. Let the surface charge density (charge per unit surface area) be σ. Let P be a point at a distance r from the sheet. We want to calculate E at P. A convenient Gaussian surface is a “pill box” of cross-sectional area A and height 2r, arranged to pierce the plane as shown. P’ is symmetrical with P, on the other side of the sheet. From symmetry, E points at right angles to the end caps and away from the plane. Also its magnitude will be same at P and P’.

. : The flux through the two plane ends is

Ф = ф E . dS + ф E . dS = ф E dS + ф E dS = EA + EA = 2EA.

The flux through the curved surface of the Gaussian cylinder is zero because E and S are at right angles everywhere on the curved surface.

. : total flux through the Gaussian cylinder = Ф = 2EA

Net charge enclosed by the Gaussian cylinder = q = σA
By Gauss’ law, 2 EA = σA/ ε0
. :        E = σ/ 2 ε0

E is independent of the distance of the point from the sheet. E is the same for all points on each side of the plane. An infinite sheet of charge cannot exist physically. However this result is true even for sheets of finite sizes if the points chosen are not near the edges and the distance r is small compared to the dimensions of the sheet.

For more help in Electric Field due to an Infinite Plan Sheet of Charge please click the button below to submit your homework assignment.